Inconsistent System of Equations? Look Closer!

Date: 11/17/2004 at 12:25:37
From: Randy
Subject: Fith Grade Story Problem that has several adults stumped.
When Alexis, Chelsea, and Kammi had lunch together, Alexis spent $1.60
for two small hamburgers, a drink and one order of fries. Chelsea's
two orders of fries, two drinks and one small burger cost $1.40
altogether. How much does Kammi owe for a small burger, one order
of fries and one drink?
It seems as though there is not enough information. I tried looking
at several of the variables and considered fractions or ratios or
percentages. However, I struggle with the fact that the prices of
each could be any non-related amount.

Date: 11/17/2004 at 12:57:14
From: Doctor Douglas
Subject: Re: Fith Grade Story Problem that has several adults stumped.
Hi Randy.
For a fifth-grader, I would do the following: put Alexis and Chelsea
together--they have a total of three hamburgers, three drinks, and
three fries for a total of $3.00. Since Kammi has one burger, one
drink, and one order of fries, she should owe one third of this, or $1.00.
At a slightly more advanced level, we would probably use algebra to
solve a problem of this type. Let H, D and F be the cost of a
hamburger, drink, and fries respectively.
Alexis: 2H + D + F = 160
Chelsea: H + 2D + 2F = 140
Kammi: H + D + F = ?
You can probably see that the shortest path to the answer to the
question is essentially encapsulated in the fifth-grade explanation
above.
Note that the question asks for the combination H + D + F, and not the
individual prices of each hamburger/drink/fries. In fact, there is
not enough information given in the problem to completely determine
the individual costs. For example, any set of {H,D,F} satisfying the
equations labelled Alexis and Chelsea will work:
H = 60, D = 30, F = 10
H = 60, D = 20, F = 20
H = 60, D = 80, F = -40 (of course this is unreasonable!)
Algebra allows us to prove that the cost of the hamburger must be
60 cents:
2*Alexis - Chelsea: 4H + 2D + 2F - H - 2D - 2F = 2*160 - 140
3H = 320 - 140 = 180
H = 60
and what remains is the relationship D + F = 160 - 2H = 40. Any
pricing combination such that the drink and the fries together add up
to 40 cents is consistent with the problem statement. All of the
solutions above are consistent with this.
I hope that this helps. Please write back if you have more questions
about this.
- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/

Date: 11/17/2004 at 13:07:28
From: Randy
Subject: Thank you (Fith Grade Story Problem that has several adults
stumped.)
Thank you Doctor Douglas for your quick response. I have a bachelors
in Business Admin but have realized how much is lost over time. I do
appreciate the services provided at this site. Now maybe I can get
caught up with my own reasoning skills and my 11 year old son might
think I actually know something!